A vibrating string of certain length ′ l ′ under a tension ′ T ′ resonates with a mode corresponding to the first overtone of an air column of length 75 cm inside a tube closed at one end. The string

Question

A vibrating string of certain length ′ l ′ under a tension ′ T ′ resonates with a mode corresponding to the first overtone of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats/s, when vibrated along with a tuning fork of frequency ′ n ′ . Increasing the tension of the string reduces the beat frequency. The frequency ′ n ′ of the tuning fork in Hz is { Velocity of sound in air is 340 m/s }

TestHub · Accepted Answer

The frequency of the first overtone for an air column inside a tube closed at one end corresponds to the third harmonic, which is given by:

$f = \frac{3 v}{4 L}$

Substituting $v = 340 m/s$ and $L = 75 cm = 0.75 m$ :

$f = \frac{3 \times 340}{4 \times 0.75} = 340 Hz$

Since the vibrating string resonates with this mode, the initial frequency of the string is $f_{s} = 340 Hz$ .

The string produces $4 beats/s$ with a tuning fork of frequency $n$ . Thus, the possible frequencies of the tuning fork are:

$n = f_{s} \pm 4$

$n = 340 + 4 = 344 Hz$ or $n = 340 - 4 = 336 Hz$

The frequency of a vibrating string is directly proportional to the square root of its tension, $f_{s} \propto T$ . Therefore, increasing the tension increases the frequency of the string ( $f_{s} > 340 Hz$ ).

Since increasing the tension reduces the beat frequency, the frequency of the string must move closer to the frequency of the tuning fork. This is only possible if the frequency of the tuning fork is greater than the initial frequency of the string ( $n > f_{s}$ ).

Hence, $n = 344 Hz$ .

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